if four out of every seven individuals in a population of armadillos carry a gene for a defective enzyme, how many individuals carry the normal gene in a population of 869 armadillos?
well try to do 869 divided by 7
(3/7) * 869 = ?
To find the number of individuals carrying the normal gene in a population of armadillos, we need to determine the proportion of individuals that do not carry the defective gene.
Given that four out of every seven individuals carry the gene for a defective enzyme, we can calculate the proportion of individuals that do not carry the defective gene by subtracting this proportion from 1.
Proportion of individuals carrying the normal gene = 1 - (4/7) = 3/7
Now, we can calculate the number of individuals carrying the normal gene by multiplying the proportion by the total population size.
Number of individuals carrying the normal gene = (3/7) * 869
To calculate this, you can divide 869 by 7 and then multiply the result by 3:
Number of individuals carrying the normal gene = (3/7) * 869 ≈ 372.43
Since we are dealing with a population of armadillos, which cannot have fractional individuals, we need to round the result to the nearest whole number.
Therefore, in a population of 869 armadillos, approximately 372 individuals carry the normal gene.